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yKpoI14uk [10]
3 years ago
5

HELP ASAP!!! 2C4H10(g) + 1302 ===> 8CO2(g) + 10H2O(g)

Chemistry
1 answer:
Ivan3 years ago
6 0

Answer:

30.34g (corrected to 4 significant figures).

Explanation:

Take the atomic mass of C=12.0, H=1.0, O=16.0.

no. of moles = mass / molar mass

So, no. of moles of butane reacted = 10 / (12x4 + 1x10)

= 0.172414 mol

Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.

From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,

meaning 1 mole of butane gives 4 moles of CO2.

Using this ratio,

we can deduce that the no. of moles of CO2 produced =  0.172414 x 4

=0.689655 mol

As mass = no. of moles x molar mass

mass of CO2 produced =   0.689655 x (12.0+16.0x2)

=30.34g (corrected to 4 significant figures).

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Answer:

The pattern of change in the electron configuration increases from left to right across the period.

Explanation:

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5 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

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3 years ago
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3 years ago
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3 years ago
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Answer:

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Explanation:

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